Problem 85 - A Complete Resource Book in Chemistry for JEE Main

Question

Equilibrium constant for the reaction \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) is 81. If the velocity constant of the forward reaction is \(162 \mathrm{~L}\) \(\mathrm{mol}^{-1} \mathrm{sec}^{-1}\), what is the velocity constant (in \(\mathrm{L} \mathrm{mol}^{-1}\) \(\mathrm{sec}^{-1}\) ) for the backward reaction? (a) 13122 (b) 2 (c) 261 (d) 243

Step-by-Step Solution

Answer

The velocity constant for the backward reaction is 2 L mol⁻¹ sec⁻¹.

Step 1: Understanding the Relationship

The equilibrium constant, denoted as \( K_c \), is a ratio of the rate constants of the forward \( k_f \) and backward \( k_b \) reactions. This relationship is expressed as \( K_c = \frac{k_f}{k_b} \). Here, \( K_c = 81 \) and \( k_f = 162 \, \mathrm{L \, mol^{-1} \, sec^{-1}} \) is given.

Step 2: Solving for the Backward Reaction Rate Constant

To find the backward rate constant \( k_b \), rearrange the equation: \( k_b = \frac{k_f}{K_c} \). Substitute the known values to get \( k_b = \frac{162}{81} \).

Step 3: Performing the Calculation

Now, calculate \( k_b = \frac{162}{81} = 2 \, \mathrm{L \, mol^{-1} \, sec^{-1}} \).

Key Concepts

Equilibrium Constant

In chemistry, the equilibrium constant, denoted as \(K_c\), is a vital concept when studying reactions at equilibrium. It is the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients in the balanced chemical equation. This value provides insight into the position of equilibrium:

• If \(K_c\) is much greater than 1, the equilibrium position favors the products, indicating a greater concentration of products compared to reactants.
• If \(K_c\) is much less than 1, the equilibrium position favors the reactants.
• If \(K_c\) is around 1, significant quantities of both reactants and products are present.

For the reaction \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\), the equilibrium constant is given as 81. This indicates a strong tendency towards the formation of products (\(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\)) at equilibrium.

Reaction Rate

The reaction rate is a measure of how quickly the reactants are converted into products in a chemical reaction. It is influenced by several factors including temperature, pressure, and concentration of reactants. In kinetic studies, it is important to note:

• The rate of a reaction provides insight into the approximate time required for the reactants to form products under specified conditions.
• Higher rates imply faster reactions, while lower rates suggest slower reactions.
• Reaction rates are influenced by the specific rate constants for both the forward and backward reactions.

In our example reaction, given that the velocity constant (or rate constant) for the forward reaction \(k_f\) is 162 \(\mathrm{L \, mol^{-1} \, sec^{-1}}\), this suggests a relatively quick conversion from reactants to products in the forward direction.

Forward and Backward Reaction Rates

Chemical reactions are dynamic processes where forward and backward reactions occur simultaneously until equilibrium is achieved. The forward reaction rate refers to the speed at which reactants are transformed into products. Conversely, the backward reaction rate measures how fast the products revert to reactants.

• The forward reaction rate is governed by the rate constant \(k_f\), which, for the given reaction, is 162 \(\mathrm{L \, mol^{-1} \, sec^{-1}}\).
• The backward reaction rate is described by the rate constant \(k_b\), which can be calculated using the equilibrium constant relation: \[K_c = \frac{k_f}{k_b}\].

Given \(K_c = 81\) and \(k_f = 162\), the backward rate constant can be determined as follows:
Rearranging the equation: \(k_b = \frac{k_f}{K_c} = \frac{162}{81} = 2\).
This simple calculation shows that the backward reaction is much slower, with \(k_b\) being 2 \(\mathrm{L \, mol^{-1} \, sec^{-1}}\), compared to the forward reaction.